Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
void minn(T &A, T B) {
  A = min(A, B);
}
template <class T>
void maxx(T &A, T B) {
  A = max(A, B);
}
const long long mod = 1e9 + 7;
struct node {
  int a[11];
  node() {
    for (int i = 1; i <= 10; ++i) a[i] = 1e9;
  }
  void push(int val) {
    if (a[10] < val) return;
    a[10] = val;
    sort(a + 1, a + 1 + 10);
  }
} f[18][100005];
int n, m, k, jump[17][100005], q, u, v, d[100005];
vector<int> a[100005];
node gop(node u, node v) {
  node res = u;
  for (int i = 1; i <= 10; ++i) res.push(v.a[i]);
  return res;
}
void DFS(int u) {
  for (int i = 1; 1 << i <= n; ++i) {
    jump[i][u] = jump[i - 1][jump[i - 1][u]];
    f[i][u] = gop(f[i - 1][u], f[i - 1][jump[i - 1][u]]);
  }
  for (int v : a[u])
    if (v != jump[0][u]) {
      jump[0][v] = u;
      d[v] = d[u] + 1;
      DFS(v);
    }
}
int lca(int u, int v) {
  if (d[u] < d[v]) swap(u, v);
  int diff = d[u] - d[v];
  for (int i = 16; i >= 0; --i)
    if (diff & (1 << i)) u = jump[i][u];
  if (u == v) return u;
  for (int i = 16; i >= 0; --i)
    if (jump[i][u] != jump[i][v]) {
      u = jump[i][u];
      v = jump[i][v];
    }
  return jump[0][u];
}
node get(int u, int k) {
  node res;
  for (int i = 0; i <= 16; ++i)
    if (k & (1 << i)) {
      res = gop(res, f[i][u]);
      u = jump[i][u];
    }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    cin >> k;
    f[0][k].push(i);
  }
  DFS(1);
  while (q--) {
    cin >> u >> v >> k;
    int p = lca(u, v);
    node x = get(u, d[u] - d[p]);
    node y = get(v, d[v] - d[p] + 1);
    node res = gop(x, y);
    vector<int> ans;
    for (int i = 1; i <= k; ++i) {
      if (res.a[i] == 1e9) break;
      ans.push_back(res.a[i]);
    }
    cout << (int)((ans).size()) << " ";
    for (int i : ans) cout << i << " ";
    cout << '\n';
  }
}